Dispersion relation of water waves

Remarks

The proof looks for a potential of the form $\phi = A(z)\sin(kx - \omega t)$. Laplace's equation and the bottom condition force $A(z) = B\cosh(k(z + h))$. The two surface conditions then have a nonzero solution only under the stated relation. In deep water, $kh \to \infty$, the relation becomes $\omega^2 = gk$, so $c = \sqrt{g/k}$ and $c_g = c/2$. In shallow water, $kh \to 0$, it becomes $c = c_g = \sqrt{gh}$, independent of $k$, so long waves do not disperse. With surface tension in deep water, $c^2 = g/k + \sigma k/\rho$ has a minimum at $k = \sqrt{\rho g / \sigma}$; for clean water this wavelength is about $1.7$ cm.

Going deeper

For a student

The relation is the basic input of linear water wave theory. Because $c_g \neq c$, the energy of a wave group travels at a different speed from the crests, and in deep water the crests run through the group at twice its speed. This fact explains the Kelvin wake pattern behind a ship. The relation also controls the weakly nonlinear theories: expanding $\tanh(kh)$ for small $kh$ gives the dispersive terms of the Boussinesq and Korteweg–de Vries equations.